Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3121 Accepted Submission(s):
778
Problem Description
Assume that f(0) = 1 and 0^0=1. f(n) = (n%10)^f(n/10)
for all n bigger than zero. Please calculate f(n)%m. (2 ≤ n , m ≤ 10^9, x^y
means the y th power of x).
Input
The first line contains a single positive integer T.
which is the number of test cases. T lines follows.Each case consists of one
line containing two positive integers n and m.
Output
One integer indicating the value of f(n)%m.
Sample Input
2
24 20
25 20
Sample Output
16
5
Source
Recommend
$a^x \equiv a^{x \ \% \ phi(m) + phi(m)} \pmod m$
然后直接上就行了。
有很多奇怪的边界问题，比如求$f(n)$的时候一模就炸。。
#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #define int long long using namespace std; const int MAXN = 1e5 + 10, INF = 1e9 + 10; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N, M, PhiM; int fastpow(int a, int p, int mod) { if(a == 0) return p == 0; int base = 1; while(p) { if(p & 1) base = (base * a) % mod; p >>= 1; a = (a * a) % mod; } return base == 0 ? mod : (base + mod)% mod; } int GetPhi(int x) { int limit = sqrt(x), ans = x; for(int i = 2; i <= limit; i++) { if(!(x % i)) ans = ans / i * (i - 1) ; while(!(x % i)) x /= i; } if(x > 1) ans = ans / x * (x - 1); return ans; } int F(int N, int mod) { if(N < 10) return N; return fastpow((N % 10), F(N / 10, GetPhi(mod)), mod); } main() { int QwQ = read(); while(QwQ--) { N = read(); M = read(); printf("%I64d\n", F(N, M)); } return 0; } /* 4 24 20 37 25 123456 321654 123456789 456789321 */
